Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 12 / c

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- Solution c

Solution

Use the embedded surface parametrization

X (r, θ) = (r cos θ, r sin θ, a (1 - \frac{r ^{2}}{b ^{2}})) .

Its tangent vectors have cross product

X_{r} \times X_{θ} = (\frac{2 a r ^{2}}{b ^{2}} cos θ, \frac{2 a r ^{2}}{b ^{2}} sin θ, r),

whose magnitude is

r 1 + 4 a^{2} r^{2} / b^{4}

. Thus the surface area is

A = α \int_{0}^{R} r 1 + \frac{4 a ^{2} r ^{2}}{b ^{4}} d r .

With

u = 1 + 4 a^{2} r^{2} / b^{4}

, the radial integral becomes

b^{4} (8 a^{2})^{- 1} \int u^{1/2} d u

. Evaluating its endpoints gives

A = \frac{α b ^{4}}{12 a ^{2}} [(1 + \frac{4 a ^{2}}{b ^{2}})^{3/2} - 1] .

This is the scalar-area part of the paraboloid sector surface measures.

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